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jeka94
2 years ago
12

An ideal solution consisting of 79 wt% benzene (C6H6) and 21 wt% toluene (C7H8) was heated in a closed vessel to 50 ºC. What is

the mole fraction of benzene in the vapour phase when equilibrium was reached at 50 ºC? Round your answer to two significant figures.
Data: Vapour pressure of benzene at 50 ºC = 271 mmHg

Vapour pressure of toluene at 50 ºC = 91.5 mmHg
Chemistry
1 answer:
Thepotemich [5.8K]2 years ago
5 0

In this case, according to the Raoult's law, we can find the mole fraction of 79 wt% benzene and 21 wt% toluene at 50 °C in the vapor phase as follows:

y_iP=x_iP_i

However, we first have to calculate the mole fractions in the solution as follows (b stands for benzene and t for toluene):

\\x_b=0.79\frac{g\ benzene}{g\ solution} *\frac{(78.11*x_b+92.14*x_t)}{1mol\ solution} *\frac{1mol\ toluene}{78.11g\ toluene} \\\\x_b=0.79\frac{g\ benzene}{g\ solution} *\frac{78.11*x_b+92.14*(1-x_b)}{1mol\ solution} *\frac{1mol\ toluene}{78.11g\ toluene}\\\\x_b=0.79*(x_b+1.18*(1-x_b))\\\\x_b=0.79x_b+0.932-0.932x_b\\\\x_b=\frac{0.932}{1+0.932-0.79} =0.816\\\\x_t=1-x_b=1-0.816=0.184

Next, we calculate the total pressure as follows, according to the Dalton's law:

P=x_bP_b+x_tP_t=0.816*271mmHg+0.184*91.5mmHg=237.972mmHg

Finally, the mole fractions of the vapor phase turn out:

y_b=\frac{0.816*271mmHg}{237.972mmHg}=0.929\\\\y_t=1- 0.929=0.071

Learn more:

  • brainly.com/question/12718562
  • brainly.com/question/17581597

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Salicylic acid (C7H6O3) reacts with acetic anhydride (C4H6O3) to form acetylsalicylic acid (C9H8O4).
zavuch27 [327]

Answer : The correct option is, (B) Salicylic acid

Solution :

First we have to calculate the moles of salicylic acid and acetic anhydride.

\text{Moles of }C_7H_6O_3=\frac{\text{Mass of }C_7H_6O_3}{\text{Molar mass of }C_7H_6O_3}=\frac{70g}{138.121g/mole}=0.507moles

\text{Moles of }C_4H_6O_3=\frac{\text{Mass of }C_4H_6O_3}{\text{Molar mass of }C_4H_6O_3}=\frac{80g}{102.09g/mole}=0.783moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2C_7H_6O_3(aq)+C_4H_6O_3(aq)\rightarrow 2C_9H_8O_4(aq)+H_2O(l)

From the balanced reaction we conclude that

As, 2 moles of salicylic acid react with 1 mole of acetic anhydride

So, 0.507 moles of salicylic acid react with \frac{0.507}{2}=0.2535 mole of acetic anhydride

The excess of acetic anhydride = 0.783 - 0.2535 = 0.5295 moles

That means the in the given balanced reaction, salicylic acid is a limiting reagent because it limits the formation of products and acetic anhydride is an excess reagent.

Hence, the limiting reagent is, salicylic acid.

7 0
2 years ago
According to the following reaction, how many grams of potassium sulfate will be formed upon the complete reaction of 23.8 grams
Alexxandr [17]

<u>Answer:</u> The mass of potassium sulfate that can be produced is 73.88 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For KOH:</u>

Given mass of KOH = 23.8 g

Molar mass of KOH = 56.1 g/mol

Putting values in equation 1, we get:

\text{Moles of KOH}=\frac{23.8g}{56.1g/mol}=0.424mol

The chemical equation for the reaction of KOH and potassium hydrogen sulfate follows:

KHSO_4+KOH\rightarrow K_2SO_4+H_2O

As, potassium hydrogen sulfate is present in excess. It is considered as an excess reagent.

KOH is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of KOH produces 1 mole of potassium sulfate

So, 0.424 moles of KOH will produce = \frac{1}{1}\times 0.424=0.424moles of potassium sulfate

Now, calculating the mass of potassium sulfate from equation 1, we get:

Molar mass of potassium sulfate = 174.26 g/mol

Moles of potassium sulfate = 0.424 moles

Putting values in equation 1, we get:

0.424mol=\frac{\text{Mass of potassium sulfate}}{174.26g/mol}\\\\\text{Mass of potassium sulfate}=(0.424mol\times 174.26g/mol)=73.88g

Hence, the mass of potassium sulfate that can be produced is 73.88 grams

8 0
3 years ago
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